Show that $\int_0^\infty P(Y < -y)dy = -\int_{-\infty}^0 x f_Y(x)dx$

Question

I am trying to understand why $\int_0^\infty P(Y < -y)dy = -\int_{-\infty}^0 x f_Y(x)dx$

My book (Ross) gives a solution where too many steps are skipped for me to follow.

Answer 1

$P(Y<x)=\int_{-\infty}^x f_Y(t)dt$. Hence, using integration by parts, \begin{align} -\int_{-\infty}^0 x f_Y(x)\,dx=-\left(x\int_{-\infty}^x f_Y(t)dt\right)\Big|_{-\infty}^0+\int_{-\infty}^0\int_{-\infty}^x f_Y(t)\,dt\,dx. \end{align} The first term vanishes and the second term becomes \begin{align} \int_{-\infty}^0 P(Y<x)\,dx. \end{align} Let $y=-x$.